Vs is and alternating voltage - this so as to setup a flux in the core.
This can be seen from Faraday's law/ Where e is the voltage set up by the varying flux.
Now we know that the flux varies as a sinusoid.
Therefore = maxsin(t)Therefore /
From the above analysis we can see that the voltage lags the flux by 90 degrees
A non-ideal transformer has eddy current losses in the core, and resistive losses in its primary and secondary windings. In addition the core of the transformer requires some finite mmf for its magnetization. Also not all the flux links the primary and secondary windings, in other words there exists leakage flux in the windings.
Using all this information on the no ideal transformer we could develop the equivalent circuit shown below for it.
In the above diagram the circuit components used are used to represent the various losses stated above. The relationship between circuit component and loss is detailed below.
Ro => This resistor represents the hysteresis and eddy-current losses. These losses are directly dependent on the input voltage, as can be seen by its position.
Xm => This inductor represents the magnetization reactance - that minimum mmf required for the magnetization of the transformer core.
R1 and R2 - These resistors represent the resistive losses (I2R) present in the windings of the primary and secondary sides of the transformer.
X1 and X2 - These represent the leakage reactance arising from leakage fluxes in the primary and secondary windings.
On No Load conditions we have the familiar relations
To make the equivalent circuit of the transformer a little simpler we can merge the resistive and inductive components of the windings as shown in the diagram below:
In the diagram shown I => current required to magnetize the core (flux per turn)
So for N1 turns on the primary side; total flux = N1
as shown previously (t) = maxsint
Therefore d/dt = maxcost
V1 = N1maxcost
Using the equivalent circuit we can see that the load losses in the transformer can be given by
I12R1 + I22R2
General points to remember: / when going to the HV side, impedances go up
when going to the LV side impedances go down
Any side can be treated as the primary, all the must be done is that the turns ratio should be adjusted to suit.
Paralleling of transformers
Say we had a situation as depicted such. A plant demanded a load of 250kVA. To account for this load the electricity commission installed a 300kVA transformer at the location of the plant, which served well to supply power to the plant. But say in the future the planted decided to expand and logically it demanded a greater load from the lines, say 500kVA. The electricity commission has two choices on how to deal with the plant. It could replace the old transformer with a new one capable of handling the new demand. Or it could add another transformer, which would be able to comfortable handle the extra 200kVA, demanded.
The second option is the more feasible one as a smaller transformer needs to be need be purchased which is much cheaper than the purchase of a brand new big transformer.
In paralleling transformers, however, some basic rules need to be followed.
Obviously the voltage and turns ratio need to be the same otherwise what would occur is that one transformer (the one with the higher secondary voltage) would be driving current into the other transformer. Similar R / X ratio in relation to the transformer size must be observed. And the polarity must be the same for both. The diagram below shows the effect of ignoring polarity.
In practice transformers come with various voltage turn ratios and various VA ratings, and these transformers would at times need to be used in the same system. When dealing with situations such as these, analyzing impedance values is problematic. What is done is that a base VA rating, and voltage are used to which all values on all the transformers are referred. The base VA rating is defined as sb, and is expressed in per unit (pu).
So Vb x Ib = sb
From this equation we can get the following relationships.
Zb = Vb / Ib
Ib = sb / Vb
Impedance of T/F pu = Actual Impedance / Base Impedance
For example say we had a 10MVA transformer with turns ratio 66/11 on an 11kV system.
Then Zb = Vb2 / sb = 11x103 / 10x103 = 12.1
Actual Z = 1.21 (say this were given),
Then pu Z = 1.21 / 12.1 = 0.1pu